Problem Statement

These define three classifications of positive integers based on their   proper divisors. Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

6   has proper divisors of   1,   2,   and   3. 1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes. Show the results here.

Let's start with the solution:

fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) }
 
fcn classify(n){
   p:=properDivs(n).sum();
   return(if(p
}
 
const rangeMax=20_000;
classified:=[1..rangeMax].apply(classify);
perfect   :=classified.filter('==(0)).len();
abundant  :=classified.filter('==(1)).len();
println("Deficient=%d, perfect=%d, abundant=%d".fmt(
   classified.len()-perfect-abundant, perfect, abundant));